If \(A = \begin{bmatrix}2&1\\1&2\end{bmatrix}\) and \(A^4 = pA + qI\), where I is the identity matrix of order 2 and p, q are real numbers, then the value of \(p - q\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 79
Let us first write \(A^2\) in terms of A and I:
\( \displaystyle A^2 = \begin{bmatrix}4 + 1&2 + 2\\2 + 2&1 + 4\end{bmatrix} = \begin{bmatrix}5&4\\4&5\end{bmatrix} \)
\( \displaystyle 4A - 3I = \begin{bmatrix}8 - 3&4\\4&8 - 3\end{bmatrix} = \begin{bmatrix}5&4\\4&5\end{bmatrix} \)
So the key relation is
\(A^2 = 4A - 3I\)
Squaring it (A and I commute):
\(A^4 = (4A - 3I)^2 = 16A^2 - 24A + 9I\)
\(= 16(4A - 3I) - 24A + 9I\)
\(= 40A - 39I\)
Since A is not a multiple of I, p and q are fixed:
\(p = 40,\quad q = -39\)
\(p - q = 40 + 39 = 79\)
Check: \(A^4 = (A^2)^2 = \begin{bmatrix}41&40\\40&41\end{bmatrix}\), and \(40A - 39I\) has elements \(80 - 39 = 41\) and 40.